BOF's probability problem...

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BOF
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BOF's probability problem...

Post by BOF » Mon Jan 27, 2014 7:26 pm

Dear All,
many years ago (quite a big many), I set a probability teaser in one of the early BDGA newsletters.
I cannot find any copy of my original working, either on computers or on paper!
If anyone has an archive of my original problem and the solution, I would be very grateful.

The detail of the problem was thus:

At the tee, X players flip discs to decide the order of play.
On each flip, if a single disc lands different to all the others, that player is 'out' i.e. next in order of play.
If no single disc lands differently (i.e. they all land the same, or a split of, for example, three 'up' and two 'down') then discs are re-flipped.
When down to the last 2 players, one will call 'odd' (discs land differently) or 'even' (discs land the same) to decide the last positions.

Question:
What is the probability, given X players at the tee, that only X number of flips are required- i.e. a player is 'out' on each flip.
So for 4 players only 4 flips are required (including the final odd/even flip).

Answers, on a virtual postacrd, to the forum...

Archive material (if you have any) to me, please.

Many thanks in advance.

BOF

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Re: BOF's probability problem...

Post by Mark.A.D » Mon Jan 27, 2014 7:37 pm

Surely if there are X players, only X-1 flips would be required as the final flip would decide the fate of 2 players
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Re: BOF's probability problem...

Post by Paul Holden » Mon Jan 27, 2014 8:01 pm

Always answer the question set in the paper Mark.
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Re: BOF's probability problem...

Post by Paul Holden » Mon Jan 27, 2014 8:28 pm

Joking aside I think it goes something like this...

The odds that for any one flip between Y players all but one disc land the same way up (assuming an equal chance for each way up) is:
(1/2)^(Y-1) , where I am using ^ to represent "to the power of"

Beginning with X players, and only looking at the flips required whilst there are 3 or more players left, because the last flip will always decide the last two players, the cumulative odds are:

(1/2)^(X-1+... ...+X-n) ,X > 2 ,n = X-2

Evaluating the sum of the arithmetic series in the exponent gives us:

(n/2)(2X+(n-1)(-1))

(n/2)(2X+1-n)

Substituting for n gives us:

((X-2)/2)(2X+1-X+2)

((X-2)/2)(X+3)

So the final odds that the minimum number of throws, which Mark correctly identifies as X-1, would be used is:

(1/2)^((X-2)/2)(X+3)

Am I even close???
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Re: BOF's probability problem...

Post by Paul Holden » Mon Jan 27, 2014 8:37 pm

On the archive front the only BDGA newsletter I have is Issue 1 and it isn't in there :) I have nothing between that and Inflight 2008, nice pic of Chris OB on the front.
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Re: BOF's probability problem...

Post by BOF » Mon Jan 27, 2014 11:18 pm

nice try, Paul, but I think it is simpler than that - unless your solution simplifies down to what I have in front of me here...

I have a nice solution just worked out, but am willing to give others a look before I reveal my solution (in case I'm right or wrong!).

I'll PM you my solution.

BOF

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Re: BOF's probability problem...

Post by BOF » Tue Jan 28, 2014 12:11 am

If only we could use the superscript BBCode -

Admin, if you feel like adding superscript and subscript functionality...

http://www.phpbbdevelopers.net/archive/ ... ?f=27&t=93

Thanks

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Re: BOF's probability problem...

Post by LostMeow » Tue Jan 28, 2014 2:05 pm

Some of this maths is beyond me, but Paul, is your initial probability formula correct?

i.e. the probability that one of the discs is different being (1/2)^(Y-1)

I don't know what the answer is, but if, say, three discs were flipped, there would be 6 ways out of 8 for one of the discs to be different, which is a probability of 3/4, but your formula would calculate it as 1/4.

The number of ways of one disc being different should be the 2nd number from each end of Pascal's triangle added together.
Then the probability is that number divided by the total in that row of the triangle, so for:

2 discs = 2/4 = 1/2
3 discs = 6/8 = 3/4
4 discs = 8/16 = 1/2
5 discs = 10/32 = 5/16
6 discs = 12/64 = 3/16
etc...

I'm not sure what formula gives that series though...
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Re: BOF's probability problem...

Post by bruce » Tue Jan 28, 2014 2:19 pm

LostMeow wrote: 2 discs = 2/4 = 1/2
3 discs = 6/8 = 3/4
4 discs = 8/16 = 1/2
5 discs = 10/32 = 5/16
6 discs = 12/64 = 3/16
etc...

I'm not sure what formula gives that series though...
At a glance: 2X/(2^X)
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Re: BOF's probability problem...

Post by LostMeow » Tue Jan 28, 2014 2:36 pm

bruce wrote:
At a glance: 2X/(2^X)
D'oh. Knew it would be deceptively simple! Anyway, is that the right probability, and can it be arithmetically progressed like Paul did with the other one?
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Re: BOF's probability problem...

Post by Paul Holden » Tue Jan 28, 2014 5:22 pm

LostMeow wrote:Some of this maths is beyond me, but Paul, is your initial probability formula correct?

i.e. the probability that one of the discs is different being (1/2)^(Y-1)

I don't know what the answer is, but if, say, three discs were flipped, there would be 6 ways out of 8 for one of the discs to be different, which is a probability of 3/4, but your formula would calculate it as 1/4.
Yeah, ignore that, I am home ill today and I think it was showing... :(
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Re: BOF's probability problem...

Post by Paul Holden » Tue Jan 28, 2014 5:36 pm

I am out by factor of Y...

(1/2)^(Y-1)

((1/2)^(-1))((1/2)^(Y))

2((1/2)^(Y)

2/(2^Y)

Whereas Bruce has:

2Y/(2^Y)
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Re: BOF's probability problem...

Post by LostMeow » Tue Jan 28, 2014 6:45 pm

Is this the answer?

http://www.wolframalpha.com/input/?i=pr ... rom+3+to+n

Formula is:

(2^(-0.5(n-1)n))n!
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Re: BOF's probability problem...

Post by Paul Holden » Tue Jan 28, 2014 6:59 pm

Looks good, I had just got to:

N! ( (1/2)^( (1/2)((N^2)-N) )

Which I think is the same?
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Re: BOF's probability problem...

Post by BOF » Thu Jan 30, 2014 5:42 pm

LostMeow, Paul - well done!

I wrote it as:

n!/(2^(0.5n(n-1))

Ten points to Gryffindor!

Class dismissed.

BOF

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